[x-∫[x:0]e^-t平方 dt]/(x^2)*sinx不好意思x趋于0

[x-∫[x:0]e^-t平方 dt]/(x^2)*sinx不好意思x趋于0 ∫(0,x)(1-e^-t^2)dt/x^3 x-->0 lim(∫[0,x](e^t^2)dt)^2/(∫[0,x](te^2t^2)dt)RT 设f(x)=∫(1,x^2) e^(-t)/t dt,求∫(0,1)xf（x）dt lim(n,0)x/(1-e^x^2)∫(0,x)e^t^2dt 设dy/dx∫(0,e^-x)f(t)dt=e^x,f(x)=? 数学φ(x)=∫(0~2x)t(e^t)dt…求φ'(x) 求limx->o(∫（0,x)e^t^2dt)^2/∫（0,x)te^2t^2dt 求limx->o(∫（0,x)e^t^2dt)^2/∫（0,x)te^2t^2dt f(x)=e^x+∫(x,0) t f(t) dt - x ∫(x,o) f(t) dt,求f(x) ∫(0,x)f(t-x)dt=e^(-x²)+1 求f(x)急. ∫(0,x) f(x-t)dt lim(x→0)〖(∫[cosx,0](e^t-t)dt/x^4 〗 lim(x→0)[∫(x,0)e^t²dt]/x 求极限x趋向于0,∫(0,x)(1-e^-t^2)dt/x^3 lim(x→0)[∫(0,x)(e^(t^2)-1)dt]/x^3 lim(x→0)[∫(0,x)(e^(t^2)-1)dt]/x^3 f(x)=x^2+∫[0~x]e^（x-t)f '(t)dt 怎么变到 f '(x)=2x+f '(x)+∫[0~x]e^（x-t)f '(t)dt